Construct a plane parallel to a line and measure their separation

$l:\mathbf r=(a,-2,-5)+\lambda(1,1,5)$. Since $(a,-2,-5)$ lies on p1, $3a-4+5=4$, so a=1.

With A=(1,−2,−5) on l, $EA=(3,3,-13)$. A normal is $(1,1,5)\times(3,3,-13)=(-28,28,0)=-28(1,-1,0)$. Using E, $\mathbf r\cdot(1,-1,0)=3$, hence p2:$x-y=3$.

$EN:\mathbf r=(-2,-5,8)+\mu(3,2,-1)$. In p1, $3(-2+3\mu)+2(-5+2\mu)-(8-\mu)=4$. Thus $-6+9\mu-10+4\mu-8+\mu=4$, $14\mu=28$, μ=2 and N=(4,−1,6).

Because p3 contains EN and must be parallel to l without containing it, take its normal as $(1,1,5)\times(3,2,-1)=(-11,16,-1)$. Through E, $\mathbf r\cdot(-11,16,-1)=22-80-8=-66$, equivalently p3:$11x-16y+z=66$.

Use A on l and E on p3. Their difference is $EA=(3,3,-13)$. The separation is

$\begin{aligned}&\frac{|EA\cdot(11,-16,1)|}{\sqrt{11^2+16^2+1^2}}\\ & =\frac{|33-48-13|}{\sqrt{378}}\\ & =\frac{28}{\sqrt{378}}.\end{aligned}$

It is nonzero, confirming l is parallel to but not contained in p3.